1. **State the problem:** We need to find the surface area of the right triangular prism MNOPQR. 2. **Identify given dimensions:** The triangular base has sides MN = 14 in, PR = 17.5 in, and OR = 20 in. The prism extends along the length corresponding to the edge OR = 20 in. 3. **Understand the shape:** The prism has two triangular bases (MNOP and PQR) and three rectangular faces connecting corresponding sides. 4. **Calculate the area of the triangular base:** Since the triangle is right-angled at N, sides MN and NO are perpendicular. We know MN = 14 in and NO = 20 in (since OR = 20 in is the prism length, NO is the other leg of the triangle). The area of the triangle is $$\text{Area}_{\triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 20 = 140$$ 5. **Calculate the perimeter of the triangular base:** The sides are MN = 14, NO = 20, and hypotenuse MO calculated by Pythagoras: $$MO = \sqrt{14^2 + 20^2} = \sqrt{196 + 400} = \sqrt{596} \approx 24.4$$ Perimeter: $$P = 14 + 20 + 24.4 = 58.4$$ 6. **Calculate the lateral surface area:** Multiply the perimeter of the base by the prism length (height) which is 17.5 in: $$\text{Lateral area} = P \times 17.5 = 58.4 \times 17.5 = 1022$$ 7. **Calculate total surface area:** Sum of lateral area plus twice the base area: $$\text{Surface area} = 2 \times 140 + 1022 = 280 + 1022 = 1302$$ 8. **Round the answer:** The surface area is approximately 1302 square inches. **Final answer:** $$\boxed{1302 \text{ square inches}}$$